How Do You Determine the Radius and Interval of Convergence for a Power Series?

camel-man
Messages
76
Reaction score
0
I need help finishing this problem I am stuck.

find radius of conv. and interval of convergence of the series. Ʃ k=0->∞ (1/k+1) (x)^k

I have found all the way up to row=1 there for it is between (0,∞) so now that means that if absolute value of x<1 it converges if >1 it diverges but I forgot how I find what x is. I thought it was R which would be 1. Someone please explain to me as elementary as possible because I am no math major this is my last course/test and I will never use this again.
 
Physics news on Phys.org
camel-man said:
I need help finishing this problem I am stuck.

find radius of conv. and interval of convergence of the series. Ʃ k=0->∞ (1/k+1) (x)^k

I have found all the way up to row=1 there for it is between (0,∞) so now that means that if absolute value of x<1 it converges if >1 it diverges but I forgot how I find what x is. I thought it was R which would be 1. Someone please explain to me as elementary as possible because I am no math major this is my last course/test and I will never use this again.

We don't do test questions here.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
Back
Top